Answer:
Step-by-step explanation:
Given the following complex values Z₁=2(cos(π/5)+i Sin(πi/5)) And Z₂=8(cos(7π/6)+i Sin(7π/6)). We are to calculate the following complex numbers;
a) Z₁Z₂ = 2(cos(π/5)+i Sin(πi/5)) * 8(cos(7π/6)+i Sin(7π/6))
Z₁Z₂ = 18 {(cos(π/5)+i Sin(π/5))*(cos(7π/6)+i Sin(7π/6)) }
Z₁Z₂ = 18{cos(π/5)cos(7π/6) + icos(π/5)sin(7π/6)+i Sin(π/5)cos(7π/6)+i²Sin(π/5)Sin(7π/6)) }
since i² = -1
Z₁Z₂ = 18{cos(π/5)cos(7π/6) + icos(π/5)sin(7π/6)+i Sin(π/5)cos(7π/6)-Sin(π/5)Sin(7π/6)) }
Z₁Z₂ = 18{cos(π/5)cos(7π/6) -Sin(π/5)Sin(7π/6) + i(cos(π/5)sin(7π/6)+ Sin(π/5)cos(7π/6)) }
From trigonometry identity, cos(A+B) = cosAcosB - sinAsinB and sin(A+B) = sinAcosB + cosAsinB
The equation becomes
= 18{cos(π/5+7π/6) + isin(π/5+7π/6)) }
= 18{cos((6π+35π)/30) + isin(6π+35π)/30)) }
= 18{cos((41π)/30) + isin(41π)/30)) }
b) z2 value has already been given in polar form and it is equivalent to 8(cos(7pi/6)+i Sin(7pi/6))
c) for z1/z2 = 2(cos(pi/5)+i Sin(pi/5))/8(cos(7pi/6)+i Sin(7pi/6))
let A = pi/5 and B = 7pi/6
z1/z2 = 2(cos(A)+i Sin(A))/8(cos(B)+i Sin(B))
On rationalizing we will have;
= 2(cos(A)+i Sin(A))/8(cos(B)+i Sin(B)) * 8(cos(B)-i Sin(B))/8(cos(B)-i Sin(B))
= 16{cosAcosB-icosAsinB+isinAcosB-sinAsinB}/64{cos²B+sin²B}
= 16{cosAcosB-sinAsinB-i(cosAsinB-sinAcosB)}/64{cos²B+sin²B}
From trigonometry identity; cos²B+sin²B = 1
= 16{cos(A+ B)-i(sin(A+B)}/64
= 16{cos(pi/5+ 7pi/6)-i(sin(pi/5+7pi/6)}/64
= 16{ (cos 41π/30)-isin(41π/30)}/64
Z1/Z2 = (cos 41π/30)-isin(41π/30)/4
